题解 | 能量项链
能量项链
https://www.nowcoder.com/practice/565e5812eeab4d8d8449adebcb6583ef
#include <bits/stdc++.h> using namespace std; typedef long long ll; const int N = 220; const int Mod = 1e9 + 7; ll a[N]; int n; ll dp[N][N]; int main(){ scanf("%d",&n); for(int i=1;i<=n;i++){ scanf("%lld",&a[i]); a[i+n] =a[i]; } for(int len=3;len<=n+1;len++){ for(int i=1,j;(j=i+len-1)<=2*n;i++){ for(int k =i+1;k<j;k++){//分割点 dp[i][j] = max(dp[i][j],dp[i][k]+dp[k][j]+a[i]*a[k]*a[j]); } } } ll ans = 0; for(int i=1;i<=n;i++){ ans = max(ans,dp[i][n+i]); } printf("%lld\n",ans%Mod); return 0; }
聚环成链,然后区间dp
#牛客春招刷题训练营#https://www.nowcoder.com/discuss/727521113110073344